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In an effort to build a dodecahedron frame in Fusion360 I need to know some of the angles. Looking around I found out that the angle between an edge and a face on a regular dodecahedron is $121.7^\circ$ but I couldn't find the mathematical formula nor the way to calculate this angle. The formula is needed so the exact angle can be used so the simulation is precise. Can anyone help?

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    $\begingroup$ Welcome to MSE. What did you try? $\endgroup$ Dec 25, 2022 at 12:12
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    $\begingroup$ Hint: Lopping-off a corner of the dodecahedron gives a tetrahedron. So ... Consider a general tetrahedron whose edges at a vertex are $a$, $b$, $c$, bounding face-angles $A$, $B$, $C$ (w/$a$ opposite $A$, etc). The volume is given by $$V=\frac16abc\sqrt{1+2\cos A\cos B\cos C-\cos^2A-\cos^2B-\cos^2C}$$ but it's also given by $$V=\frac13\cdot(\text{area of base})\cdot\text{height}=\frac13\cdot\frac12ab\sin C\cdot c\sin\theta$$ where $\theta$ is the angle that the edge $c$ makes with the plane of the base. Equating the forms of $V$ gives $\theta$ in terms of $A, B, C$. (Lengths $a, b, c$ cancel.) $\endgroup$
    – Blue
    Dec 25, 2022 at 15:21
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    $\begingroup$ This comment isn't really mathematical per se, but if your goal is to construct a functional model, I would expect it to be much easier to use vertex positions rather than angles. $\endgroup$ Dec 26, 2022 at 18:59

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Make a small sphere, using a vertex of the dodecahedron as the centre of the sphere. The intersection of the sphere surface and the dodecahedron surface is a triangle, whose edges are circular arcs with length $108^\circ$ (the angle in a pentagon). The angle $\phi$ you seek is the altitude of this triangle; that's the length of the arc connecting a vertex of the triangle (corresponding to an edge of the dodecahedron) perpendicularly to the opposite edge of the triangle (corresponding to a face of the dodecahedron). The spherical law of cosines, applied to the triangle cut in half, gives

$$\cos\phi=\cos108^\circ\cos54^\circ+\sin108^\circ\sin54^\circ\cos\theta,$$

where $\theta$ is the dodecahedron's dihedral angle, which in turn is given by the spherical law of cosines applied to the whole triangle:

$$\cos108^\circ=\cos108^\circ\cos108^\circ+\sin108^\circ\sin108^\circ\cos\theta$$ $$\cos\theta=\frac{\cos108^\circ-\cos^2 108^\circ}{\sin^2 108^\circ}.$$

I assume you know that the regular pentagon angle has $\cos108^\circ=-\tfrac12\varphi^{-1}$ and $\sin108^\circ=\tfrac12\sqrt{\varphi\sqrt5}$, where $\varphi=\tfrac12(1+\sqrt5)$ is the golden ratio. So we get

$$\cos\theta=-\frac{\sqrt5}{5},$$ $$\cos\phi=\left(-\tfrac12\varphi^{-1}\right)\left(\tfrac12\sqrt{\varphi^{-1}\sqrt5}\right)+\left(\tfrac12\sqrt{\varphi\sqrt5}\right)\left(\tfrac12\varphi\right)\left(-\frac{\sqrt5}{5}\right)$$ $$=-\sqrt\frac{\varphi^{-1}}{\sqrt5}\quad=-\sqrt\frac{5-\sqrt5}{10}.$$

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First, define the golden ratio and its reciprocal

$$ a:=(1+\sqrt{5})/2, \quad b:=1/a=a-1. $$

From the Wikipedia article regular dodecahedron, define the six dodecahedron vertices $$ v_1 = (0,a,b),\; v_2 = (0,a,-b),\; v_3 = (1,1,-1),\\ v_4 = (a,b,0),\; v_5 = (1,1,1),\; v_6 = (a,-b,0) $$ where $\{v_1,v_2,v_3,v_4,v_5\}$ are the vertices of a pentagonal face and $v_6-v_4$ is an edge vector. Let $v_0 := (v_1+v_2)/2 = (0,a,0)$ be the midpoint of the edge of the face opposite to $v_4$. Then the acute angle $\theta$ between $w_1 := v_0-v_4 = (-a,1,0)$ and $w_2 := v_4-v_6 = (0,2b,0)$ can be computed from $$ c := \cos(\theta) = \frac{|w_1\cdot w_2|}{|w_1|\,|w_2|} = \frac{2b}{\sqrt{a^2+1}(2b)}= \frac1{\sqrt{a^2+1}} = \sqrt{\frac{5-\sqrt{5}}{10}}. $$ Note that $ c \approx 0.5257311121 $ and $\theta \approx 58.2825255^\circ$ while the angle between a face and an edge it adjoins is $180^\circ-\theta.$

Note the advantage of this method is that it needs only the coordinates of a few vertices using the golden ratio, the dot product of vectors, and no trigonometry except at the very end to get $\theta$ from its cosine.

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It is maybe not a good habit to use short-cut formulas/constructions from Wikipedia, since a real mathematician should construct his geometric or analytic geometric figures herself/himself, however, I believe that it is necessary when you have not enough time.

Wikipedia already constructed a dodec with 20 vertices: $(\pm 1,\pm 1, \pm 1), (0,\pm\phi,\pm\phi^{-1}), (\pm\phi^{-1},0,\pm\phi), (\pm\phi,\pm\phi^{-1},0).$

I considered the face given by the plane $x+\phi y=\phi^2$ which has vertices, $(1,1,1),(1,1,-1),(\phi,\frac{1}{\phi},0)$, $(0,\phi,\frac{1}{\phi}),(0,\phi,-\frac{1}{\phi})$ and with normal vector $\vec n=\langle1,\phi,0\rangle$.

An edge really connects the vertices $P=(1,1,1)$ and $Q=(\frac{1}{\phi},0,\phi)$. And by using $\frac{1}{\phi}=\phi-1$, we have $\vec{PQ}=\langle \phi-2,-1,\phi-1\rangle$.

Now, we will find the angle between $\vec n$ and $\vec{PQ}$: $$\langle1,\phi,0\rangle\cdot\langle \phi-2,-1,\phi-1\rangle=\sqrt{1+\phi^2}\sqrt{(\phi-2)^2+(-1)^2+(\phi-1)^2}\cos\theta$$ $$-2=\sqrt{1+\phi^2}\sqrt{6-6\phi+2\phi^2}\cos\theta$$ Now by using $\phi^2=\phi+1$ and $\phi=\frac{\sqrt5+1}{2}$ $$-2=2\sqrt{2+\phi}\sqrt{2-\phi}\cos\theta$$ $$-1=\sqrt{4-\phi^2}\cos\theta$$ $$\theta=\arccos\left(-\frac{1}{\sqrt{3-\phi}}\right)=\arccos\left(-\frac{\sqrt2}{\sqrt{5-\sqrt5}}\right)=\arccos\left(-\frac{\sqrt{5+\sqrt5}}{\sqrt{10}}\right)\approx 148.3^{\circ}$$

Hence, the obtuse angle between the face and the edge is $\alpha=270^{\circ}-\theta=121.7^{\circ}$.

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This angle can be derived from the dihedral angle $\delta = \arccos (-1/\sqrt{5})$ of the dodecahedron using a vector method. (The dihedral angle $\delta$ is derived during construction of the dodecahedron - eg see this answer to How does this proof of the regular dodecahedron's existence fail?).

Consider a group of three adjacent pentagon faces as shown in Fig 1, in which pentagon $A$ is in the plane of the page, and $B$ and $C$ are tilted forwards to meet $A$ at the dihedral angle $\delta$. Each pentagon makes dihedral angle $\delta$ with the other two. The angle required is $\alpha$ between edge $e$ and the plane of pentagon $A$. If $\beta$ is the angle between the unit normal $\underline{\mathbf{u}}_A = \underline{\mathbf{k}}$ to $A$ and the unit vector $\underline{\mathbf{e}}$ along $e$, then $\alpha = \beta + 90$. Since edge $e$ is the intersection line of the planes $B$ and $C$, then if $\underline{\mathbf{u}}_B$ is a unit normal to $B$ and $\underline{\mathbf{u}}_C$ is a unit normal to $C$ then noting that the angle between $\underline{\mathbf{u}}_B$ and $\underline{\mathbf{u}}_C$ is the dihedral angle $\delta$, we have : $$\underline{\mathbf{e}} = \frac{1}{|\underline{\mathbf{u}}_B \times \underline{\mathbf{u}}_C|} \; (\underline{\mathbf{u}}_B \times \underline{\mathbf{u}}_C) = \frac{1}{\sin \delta} \; (\underline{\mathbf{u}}_B \times \underline{\mathbf{u}}_C). $$


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Setting up an $xyz$-axes system as shown in Figs 1 and 2 we have : \begin{eqnarray*} \underline{\mathbf{u}}_B & = & -\sin \epsilon\; \underline{\mathbf{i}} + \cos \epsilon\; \underline{\mathbf{k}} \\ \Rightarrow \hspace{1em} \underline{\mathbf{u}}_B & = & -\sin \delta\; \underline{\mathbf{i}} - \cos \delta\; \underline{\mathbf{k}} \end{eqnarray*}

Since pentagon $C$ is obtained from pentagon $B$ by rotating $B$ by $+72^\circ$ about the $z$-axis, we can obtain $\underline{\mathbf{u}}_C$ from $\underline{\mathbf{u}}_B$ by rotating $\underline{\mathbf{u}}_B$ by $72^\circ$ about the $z$-axis. (Note we are rotating vectors, which possess only direction and magnitude - they have no starting or ending point, in contrast to directed line segments, but can be 'placed' anywhere and will always point in the same direction). This rotation is a 2D rotation in the $xy$-plane and leaves the $\underline{\mathbf{k}}$ component unchanged, resulting in : $$ \underline{\mathbf{u}}_C = -\sin \delta \cos 72\; \underline{\mathbf{i}} - \sin \delta \sin 72\; \underline{\mathbf{j}} - \cos \delta\; \underline{\mathbf{k}} $$

Then : $$ \underline{\mathbf{u}}_B \times \underline{\mathbf{u}}_C = \sin^2 \delta \sin 72\; \underline{\mathbf{k}} -\sin \delta \cos \delta\; \underline{\mathbf{j}} + \sin \delta \cos \delta \cos 72\; \underline{\mathbf{j}} - \sin \delta \cos \delta \sin 72\; \underline{\mathbf{i}} $$

and so : $$ \cos \beta = \underline{\mathbf{k}} \cdot \underline{\mathbf{e}} = \frac{1}{\sin \delta} \; (\underline{\mathbf{u}}_B \times \underline{\mathbf{u}}_C) \cdot \underline{\mathbf{k}} = \sin \delta \sin 72 $$

and the desired angle $\alpha$ is given by : $$ \cos \alpha = \cos(\beta + 90) = -\sin \beta = -\sqrt{1 - \cos^2 \beta}. $$

The 18-72-90$^\circ$ triangle (which comes from the diagonal of a regular pentagon) tells us : $$ \sin 72 = \frac{\sqrt{4\Phi + 3}}{2\Phi} $$

where $\Phi$ is the golden ratio $\frac{1 + \sqrt5}{2}$.

The dihedral angle $\delta$ satisfies $\sin^2 \delta = 4/5$ so making use of the identity $\Phi^2 = \Phi + 1$ we now have: \begin{eqnarray*} \cos^2 \beta & = & \sin^2 \delta \sin^2 72 = \frac{4}{5} \cdot \frac{4\Phi + 3}{4\Phi^2} = \frac{4\Phi + 3}{5\Phi + 5} \\ \Rightarrow 1 - \cos^2 \beta & = & \frac{\Phi + 2}{5\Phi^2} \\ \Rightarrow \cos \alpha & = & -\frac{1}{\Phi \sqrt{5}} \sqrt{\Phi + 2} = -\frac{1}{\Phi \sqrt{5}} \sqrt{\frac{5 + \sqrt{5}}{2}} = -\frac{2}{1 + \sqrt{5}} \sqrt{\frac{5 + \sqrt{5}}{10}} \\ & = & -\frac{1}{2} \sqrt{\frac{(\sqrt{5} - 1)^2 (5 + \sqrt{5})}{10}} = -\frac{1}{2} \sqrt{\frac{(6 - 2\sqrt{5}) (5 + \sqrt{5})}{10}} \\ & = & -\frac{1}{2} \sqrt{\frac{20 - 4\sqrt{5}}{10}} = -\sqrt{\frac{5 - \sqrt{5}}{10}}. \end{eqnarray*}

Thus : $$ \alpha = \arccos \left\{ -\sqrt{\frac{5 - \sqrt{5}}{10}} \right\}. $$

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